Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions: Introduction to Linear Polynomials

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions cover linear relationships from Introduction to Linear Polynomials. This exercise helps students find the values of a and b in equations of the form y = ax + b.

In Ganita Manjari Class 9 Chapter 2 Exercise 2.5, students work with monthly bills, module access, gym charges, badminton court usage and temperature conversion. These questions show how a linear relationship Class 9 can be formed when two quantities are connected by a fixed rate and a fixed starting value. The solutions below explain each answer step by step using y = ax + b Class 9 methods.

Key Takeaways

Linear Relationship: A relation of the form y = ax + b represents a straight-line pattern.
Rate of Change: The value of a shows how much y changes for each unit increase in x.
Fixed Value: The value of b shows the fixed charge or starting amount.
Two Data Points: Two given values are enough to find a and b.

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5 Solutions

Exercise Set 2.5 is based on linear relationships between two variables. A linear relationship between x and y can be written as:

y = ax + b

Here:

• a shows the constant rate of change,
• b shows the fixed value or starting amount,
• x is the input variable,
• y is the output value.

These Class 9 Maths linear relationship answers help students connect algebraic expressions Class 9 with real-life situations such as bills, fees and temperature scales.

Exercise 2.5 Question 1

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

Solution:

The relation is:

y = ax + b

Given:

• When x = 10, y = 400
• When x = 14, y = 500

Substitute the values.

For x = 10:

400 = 10a + b

For x = 14:

500 = 14a + b

Now subtract the first equation from the second equation.

500 − 400 = (14a + b) − (10a + b)

100 = 4a

a = 25

Now substitute a = 25 in:

400 = 10a + b

400 = 10(25) + b

400 = 250 + b

b = 150

So, the linear relationship is:

y = 25x + 150

Answer: a = 25 and b = 150.

Exercise 2.5 Question 2

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution:

The relation is:

y = ax + b

Given:

• When x = 10, y = 800
• When x = 15, y = 1100

Substitute the values.

For x = 10:

800 = 10a + b

For x = 15:

1100 = 15a + b

Now subtract the first equation from the second equation.

1100 − 800 = (15a + b) − (10a + b)

300 = 5a

a = 60

Now substitute a = 60 in:

800 = 10a + b

800 = 10(60) + b

800 = 600 + b

b = 200

So, the linear relationship is:

y = 60x + 200

Answer: a = 60 and b = 200.

Exercise 2.5 Question 3

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a°F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.

Solution:

The relation is:

°C = a°F + b

Use the given temperature points:

• When °F = 32, °C = 0
• When °F = 212, °C = 100

Substitute the first point.

0 = 32a + b

Substitute the second point.

100 = 212a + b

Now subtract the first equation from the second equation.

100 − 0 = (212a + b) − (32a + b)

100 = 180a

a = 100 / 180

a = 5 / 9

Now substitute a = 5 / 9 in:

0 = 32a + b

0 = 32(5 / 9) + b

0 = 160 / 9 + b

b = −160 / 9

So, the relationship is:

°C = (5 / 9)°F − 160 / 9

This can also be written as:

°C = (5 / 9)(°F − 32)

Answer: a = 5 / 9 and b = −160 / 9.

Final Answers for Exercise 2.5

Concept Used in Introduction to Linear Polynomials Exercise 2.5

Introduction to Linear Polynomials Exercise 2.5 focuses on finding a linear relationship between two quantities. A linear relationship is written as:

y = ax + b

In this form, a represents the rate at which y changes when x increases by 1. The value b represents the fixed part of the relationship.

For example, in the learning platform question:

y = 25x + 150

This means:

• ₹25 is charged per module,
• ₹150 is the fixed monthly fee.

Similarly, in the gym question:

y = 60x + 200

This means:

• ₹60 is charged per hour,
• ₹200 is the fixed monthly fee.

These ideas are important in Class 9 Ganita Manjari linear polynomials solutions because they connect linear polynomials with real-life bills, charges and conversions.

About Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5

Ganita Manjari Class 9 Chapter 2 Exercise 2.5 appears in the section on linear relationships. It comes after students learn linear patterns, linear growth and linear decay.

These Class 9 Maths Chapter 2 Exercise 2.5 Solutions prepare students for later topics such as:

• equations of the form y = ax + b,
• slope and fixed cost,
• linear relationships,
• graphing straight lines,
• linear polynomial Class 9 concepts,
• real-life algebraic modelling.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2